A 5mm high pin is placed at a distance of 15cm from a convex lens of focal length 10cm. A second lens of focal length

Q: A 5mm high pin is placed at a distance of 15cm from a convex lens of focal length 10cm. A second lens of focal length 5cm is placed 40cm from the first lens and 55cm from the pin. Find 1. The position of the final image. 2. Its nature. 3. Its size.

1. First lens: u = -15cm, f = 10cm $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}-\Big(-\frac{1}{15}\Big)=-\frac{1}{10}$ $\Rightarrow\text{v}=30\text{cm}$ So, the final image is formed 10cm right of second lens. 2. m for $1^{st}$ lens: $\frac{\text{v}}{\text{u}}=\frac{\text{h}_{\text{image}}}{\text{h}_{\text{object}}}\Rightarrow\Big(\frac{30}{-15}\Big)=\frac{\text{h}_{\text{image}}}{5{\text{mm}}}$ $\Rightarrow\text{h}_{\text{image}}=-10\text{mm}$ (inverted) Second lens: $u = -(40 - 30) = -10cm; f = 5cm$ [since, the image of $1^{st}$ lens becomes the object for the second lens] $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}-\Big(-\frac{1}{10}\Big)=\frac{1}{5}$ $\Rightarrow\text{v}=10\text{cm}$ m for $2^{nd}$ lens: $\frac{\text{v}}{\text{u}}=\frac{\text{h}_{\text{image}}}{\text{h}_{\text{object}}}\Rightarrow\Big(\frac{10}{10}\Big)=\frac{\text{h}_{\text{image}}}{{-10}}$ $\Rightarrow\text{h}_{\text{image}}=10\text{mm}$ (erect, real). 3. So, size of final image = 10mm

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A 5mm high pin is placed at a distance of 15cm from a convex lens of focal length 10cm. A second lens of focal length 5cm is placed 40cm from the first lens and 55cm from the pin. Find

The position of the final image.
Its nature.
Its size.

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Solution

First lens:

u = -15cm, f = 10cm
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}-\Big(-\frac{1}{15}\Big)=-\frac{1}{10}$
$\Rightarrow\text{v}=30\text{cm}$
So, the final image is formed 10cm right of second lens.

m for $1^{st}$ lens:

$\frac{\text{v}}{\text{u}}=\frac{\text{h}_{\text{image}}}{\text{h}_{\text{object}}}\Rightarrow\Big(\frac{30}{-15}\Big)=\frac{\text{h}_{\text{image}}}{5{\text{mm}}}$
$\Rightarrow\text{h}_{\text{image}}=-10\text{mm}$ (inverted)
Second lens:
$u = -(40 - 30) = -10cm; f = 5cm$
[since, the image of $1^{st}$ lens becomes the object for the second lens]
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow\frac{1}{\text{v}}-\Big(-\frac{1}{10}\Big)=\frac{1}{5}$
$\Rightarrow\text{v}=10\text{cm}$
m for $2^{nd}$ lens:
$\frac{\text{v}}{\text{u}}=\frac{\text{h}_{\text{image}}}{\text{h}_{\text{object}}}\Rightarrow\Big(\frac{10}{10}\Big)=\frac{\text{h}_{\text{image}}}{{-10}}$
$\Rightarrow\text{h}_{\text{image}}=10\text{mm}$ (erect, real).

So, size of final image = 10mm

STD 12 Science
Physics
Ray Optics and Optical Instruments
NCERT

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